Question

Find the quantity of heat required to convert 40 g of ice at `-20^(@)`C into water at `20^(@)`C. Given `L_(ice) = 0.336 xx 10^(6) J/kg.` Specific heat of ice = 2100 J/kg-K
Specific heat of water = 4200 J/kg-K

Answer 1

Heat required to raise the temperature of ice from `-20^(@)C` to `0^(@)C= 0.04xx2100xx20 =1680J`
Heat required to convert the ice into water at `0^(@)C` = `mL = 0.04xx0.336xx10^(6)=13440J`
Heat required to heat water from `0^(@)C` to `20^(@)C = 0.04xx4200xx20 = 3360J`
Total heat required =`1680+13440+3360 = 18480J`